A block of mass 0.90 kg rests on a rough horizontal floor (μ=0.20), attached to a light spring of constant 100 N m^-1 at its natural length. A bullet…

The embedding is a perfectly inelastic collision of very short duration, so momentum is conserved for the bullet and block: $0.10\times100=(1.0)V$, giving $V=10\ \mathrm{m\,s^{-1}}$ and kinetic energy $\tfrac12(1.0)(10)^2=50\ \mathrm{J}$ (most of the bullet's energy was lost in the collision). The combined mass of $1.0\ \mathrm{kg}$ then compresses the spring against friction $f=\mu(1.0)g=1.96\ \mathrm{N}$. At maximum compression $x$ the speed is zero, so $50=\tfrac12(100)x^2+1.96x$, i.e. $50x^2+1.96x-50=0$. Solving, $x=\dfrac{-1.96+\sqrt{1.96^2+4\cdot50\cdot50}}{100}=0.9806\ \mathrm{m}\approx0.98\ \mathrm{m}$. Ignoring friction would give $1.00\ \mathrm{m}$, and using only the block's mass $0.90\ \mathrm{kg}$ in the friction term gives a slightly different value.